Problem: Find the slope and y-intercept of the line that is ${\text{parallel}}$ to $\enspace {y = -\dfrac{2}{3}x + 5}\enspace$ and passes through the point ${(2, -5)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Answer: Parallel lines have the same slope. The slope of the blue line is ${-\dfrac{2}{3}}$ , so the equation of our parallel line will be of the form $\enspace {y = -\dfrac{2}{3}x + b}\enspace$ We can plug our point, $(2, -5)$ , into this equation to solve for ${b}$ , the y-intercept. $-5 = {-\dfrac{2}{3}}(2) + {b}$ $-5 = -\dfrac{4}{3} + {b}$ $-5 + \dfrac{4}{3} = {b} = -\dfrac{11}{3}$ The equation of the parallel line is $\enspace {y = -\dfrac{2}{3}x - \dfrac{11}{3}}\enspace$. ${m = -\dfrac{2}{3}, \enspace b = -\dfrac{11}{3}}$